An intriguing supremum… which may be inaccurate

Question

I found the following exercise and I would like to know whether the property is true or not, and above all how to prove it:

for $z$ so that $\Im z>0$,

$$\sup_{t\in\mathbb{R}}\left|\frac{t-i}{t-z}\right|=\frac{|z-i|+|z+i|}{2\Im z}.$$

I was able to prove it for $z=iy$, with $y$ a real nonnegative number, but I haven't succeeded in the general case, and I even doubt it is true.

Answer 1

Yes, the given property holds.

Let $z=a+ib$ with $b>0$ and, without loss of generality, $a>0$ (you have already solved the case $a=0$). Then $$f(t):=\left|\frac{t-i}{t-z}\right|^2=\frac{t^2+1}{(t-a)^2+b^2}.$$ By considering the derivative $f'(t)$ show that the maximum value is attained at $t_0=x_0+R$ (the minimum point is $x_0-R$), where $x_0=\frac{a^2+b^2-1}{2a}$ and $R=|x_0-i|=\sqrt{x_0^2+1}$. Note that $x_0$ is the unique point on the real line such that $|x_0-i|=|x_0-z|$. Hence the circle centered at $x_0$ and of radius $R$ passes through $\pm i$, $\pm z$, and $t_0$. Finally, verify that for such $t_0$ we have that $$\left|\frac{t_0-i}{t_0-z}\right|=\frac{|z-i|+|z+i|}{2\text{Im}(z)}.$$ In fact, $i=x_0+Re^{i\theta}$ and $z=x_0+Re^{i\phi}$, with $0<\phi<\theta<\pi$,
$$\left|\frac{t_0-i}{t_0-z}\right|=\left|\frac{x_0-i+R}{x_0-z+R}\right|=\left|\frac{1-e^{i\theta}}{1-e^{i\phi}}\right|= \frac{\sin(\theta/2)}{\sin(\phi/2)}$$ and $$\begin{align}\frac{|z-i|+|z+i|}{2\text{Im}(z)} &=\frac{|e^{i\phi}-e^{i\theta}|+|e^{i\phi}-e^{-i\theta}|}{2\text{Im}(e^{i\phi})}\\ &=\frac{\sin{((\theta-\phi)/2})+\sin{((\theta+\phi)/2})}{\sin{\phi}} \\&=\frac{\sin(\theta/2)\cos(\phi/2)}{\sin(\phi/2)\cos(\phi/2)} =\frac{\sin(\theta/2)}{\sin(\phi/2)}.\end{align}$$ and the equality is verified.

Answer 2

Just a remark concerning the proof given by Robert Z.: as it strongly uses the fact that $a\neq 0$, the (simpler) case $a=0$ is to be addressed. In that specific case $$f(t)=\frac{t^2+1}{t^2+b^2}$$ and the sign of the derivative of $f$ is given by that of $t(b^2-1)$. Hence, if $|b|<1$, the supremum is 1 and is attained at the infinite, and if $|b|\geqslant 1$, the maximum is attained at $t=0$ and its value is $\frac{1}{y^2}$. Therefore, the supremum of $f$ is equal to $\max\{1,\frac{1}{y^2}\}$. Then a simple calculation shows the required identity, since $$|z-i|+|z+i|=|b-1|+|b+1|=\left\{\begin{array}{ll} 2&\mbox{ if } b\in ]0;1[\\ 2b&\mbox{ if } b\geqslant 1 \end{array}\right.$$